Ask question

Ask question

All categories

3 years ago

5

Help! I asked one of the brainly tutors and they got it wrong! Please help!

1 answer:

sladkih [1.3K]3 years ago

5 0

Step-by-step explanation:

12x-15-4x+1=-6

12x-4x= -6+15-1

8x= 8

x=1

my answer is 100 % correct

You might be interested in

Answer those two questions please

yuradex [85]

Answer:

1. 3212 = 3.212 X 10 of 3

2. because it easier to write and read

6 0

3 years ago

A triangle has a side length of 4 centimeters and an area of 12 square centimeters. A similar triangle has a corresponding side

Setler [38]

Answer:

The area of the larger triangle is 48 square centimeters

Step-by-step explanation:

Please kindly check the attached file for details.

4 0

3 years ago

Which fraction represents the decimal 0.8888...?

LiRa [457]

Answer:

1,111/1,250

Step-by-step explanation:

4 0

3 years ago

A bag contains 3 green marbles, 7 blue marbles and 2 black marbles. The probability of randomly picking a green marble is 1/4. W

Dima020 [189]

The probability of not picking a green is 3/4. To get this answer you would add up the blue and black marbles, giving you an answer of 9. You would then put 9 over 12, which simplified would equal 3/4. Another way to find the answer is to take 1 and subtract 1/4 because 1/4 is the probability of green marbles. Hope this helps!! Have an amazing rest of your day!! :) <span />

7 0

3 years ago

Find an equation of the tangent to the curve at the given point by both eliminating the parameter and without eliminating the pa

konstantin123 [22]

Answer:

$y=2x-8$

Step-by-step explanation:

The given parametric equation is;

$x=6+ln(t),y=t^2+3$

<h3><u>BY ELIMINATING THE PARAMETER</u></h3>

To eliminate the parameter we make $t$ the subject in one equation and put it inside the other.

We make $t$ the subject in $x=6+ln(t)$ because it is easier.

$\Rightarrow x-6=ln(t)$

$\Rightarrow {e}^{x-6}=e^{ln(t)}$

$\Rightarrow {e}^{x-6}=t$

Or

$t={e}^{x-6}$

We now substitute this into $y=t^2+3$ .

This gives us;

$y=(e^{x-6})^2+3$ .

$\Rightarrow y=e^{2(x-6)}+3$ .

We have now eliminated the parameter.

The equation of the tangent at (6,4) is given by;

$y-y_1=m(x-x_1)$

where the gradient function is given by;

$\frac{dy}{dx}=2e^{2(x-6)}$

We substitute $x=6$ into the gradient function to obtain the gradient.

$\Rightarrow m=2e^{2(6-6)}$

$\Rightarrow m=2e^0$

$\Rightarrow m=2$

The equation of the tangent becomes

$y-4=2(x-6)$

We simplify to obtain

$y=2x-12+4$

$y=2x-8$

<h3><u>WITHOUT ELIMINATING THE PARAMETER</u></h3>

The given parametric equation is;

$x=6+ln(t),y=t^2+3$

For $x=6+ln(t)$

$\frac{dx}{dt}=\frac{1}{t}$

For $y=t^2+3$

$\frac{dy}{dt}=2t$

The slope is given by;

$\frac{dy}{dx}=\frac{\frac{dy}{dt} }{\frac{dx}{dt} }$

$\frac{dy}{dx}=\frac{2t }{\frac{1}{t} }$

$\frac{dy}{dx}=2t^2$

At the point, (6,4), we plug in any of the values into the parametric equation and find the corresponding value for $t$ .

Notice that

When $x=6$ , $6=6+\ln(t)$

$6-6=\ln(t)$

$0=\ln(t)$

$e^0=e^\ln(t)$

$1=t$

when $y=4$ , $4=t^2+3$

$4-3=t^2$

$1=t^2$

$t=\pm1$

But the slope is the same when we plug in any of these values for t.

$\frac{dy}{dx}=2(\pm1)^2=2$

The equation of the tangent becomes

$y-4=2(x-6)$

We simplify to obtain

$y=2x-12+4$

$y=2x-8$

7 0

3 years ago